Consider the series $$\sum_{h=1}^{\frac{k-1}{2}} \sum_{n=0} ^\infty\frac1{2n+1}\tan\left(\frac{(2n+1)h}{k}\pi\right)$$ where $k$ is an odd prime. Is it always positive?
Moreover, is there a closed expression for the finite sum $\sum_{h=1}^{\frac{k-1}{2}} \tan\frac{h\pi n}{k}$ for $k$ an odd prime?
Yes, there is a glosed form for $$f(k)=\sum_{h=1}^{\frac{k-1}{2}}\tan \left(\frac{\pi h n}{k}\right)$$ but it is in terms of the q-digamma function.
For brevity, let $x=e^{\frac{2 i \pi n}{k}}$ $$f(k)=\frac{i}{2} (k-1)+$$ $$\frac {2i}{\log(x)}\left(\psi _x^{(0)}\left(1-\frac{i \pi }{\log (x)}\right)-\psi _x^{(0)}\left(\frac{1}{2} \left(k-\frac{2 i \pi }{\log (x)}+1\right)\right)\right)$$
The fact that $k$ is an odd prime numbers does not make the problem different.